// Numbas version: exam_results_page_options {"name": "Simultaneous equations by elimination 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "ans1", "ans2", "ans3", "yCoef", "n1", "n2", "n3", "n4"], "name": "Simultaneous equations by elimination 3", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

Solve the pair of equations

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\$\\begin{eqnarray} \\simplify{{n1}*x + {n2}*y -{ans1}} = 0 &&&&&&&(1)\\\\ \\simplify{{n3}*x + {n4}*y -{ans2}} = 0 &&&&&&&(2)\\end{eqnarray}\$

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We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:

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$\\simplify{{n3}*{n1}*x + {n3}*{n2}*y +{ans1}*(-1)*{n3}} = 0$

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$\\simplify{{n1}*{n3}*x + {n1}*{n4}*y +{ans2}*(-1)*{n1}} = 0$

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We then subtract one new equation from the other to get:

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$\\simplify{{yCoef}y - {ans3} = 0}$

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Now we can work out $y$

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$y = \\var{b}$

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and substitute this value back in to any of the previous equations to get the value for $x$.

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$\\simplify{{n1}*x + {n2}*{b} + (-1)*{ans1}}$ = 0

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which then solves to give $x = \\var{a}$.

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", "rulesets": {}, "parts": [{"prompt": "

We are going to solve for $y$ first. To do this, we need to eliminate $x$ from the equations. The quickest way to do this is to multiply the first equation by the co-efficient of $x$ in the second equation (here $\\var{n3}$), and multiply the second equation by the co-efficient of $x$ in the first equation (here $\\var{n1}$). We then get the equations:

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[[1]]$\\simplify{y = {ans1}*{n3}}-$[[0]]$x$

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[[3]]$\\simplify{y = {ans2}*{n1}}-$ [[2]]$x$

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We then subtract one new equation from the other to get:

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[[4]]$\\simplify{y = {ans3}}$

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Now we can work out $y$

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$y =$[[5]]

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and substitute this value back in to any of the previous equations to get the value for $x$.

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[[6]] $=\\simplify{{ans1}-{n1}*x}$

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which then solves to give $x =$[[7]].

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Solve the pair of equations

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\$\\begin{eqnarray} \\simplify{{n2}*y = {ans1} - {n1}*x} &&&&&&&(1)\\\\ \\simplify{{n4}*y = {ans2} - {n3}*x} &&&&&&&(2)\\end{eqnarray}\$

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Straightforward solving linear equations question

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